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\(A=\frac{\text{1.5.6 + 2.10.12 + 3.15.18 + 4.20.24 + 5.25.30}}{\text{1.3.5 + 2.6.10 + 3.9.15 + 4.12.20 + 5.15.25 }}\)
\(=\frac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
\(=\frac{30.\left(1+2+3+4+5\right)}{15.\left(1+2+3+4+5\right)}\)
\(=\frac{30}{15}=2\)
Vậy A=2.
\(=\frac{1.5.6+\left(1.5.6\right).2+\left(1.5.6\right).3+\left(1.5.6\right).4+\left(1.5.6\right).5}{1.3.5+\left(1.3.5\right).2+\left(1.3.5\right).3+\left(1.3.5\right).4+\left(1.3.5\right).5}\)
\(=\frac{\left(1.5.6\right).\left(1+2+3+4+5\right)}{\left(1.3.5\right).\left(1+2+3+4+5\right)}=\frac{1.5.6}{1.3.5}=\frac{1.1.2}{1.1.1}=2\)
A= \(\frac{1.5.3.2+2.10.2.6+2.15.9.2+4.20.12.2+5.25.15.2}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)
A= \(\frac{2+2+2\cdot2+2+2}{0+0+3+0+0}\)
A= \(\frac{12}{3}\)
A= 4
Đầu tiên bạn tách ra, rút gọn rồi cộng lại,tính nha!
A=\(\dfrac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
A=\(\dfrac{\left(1+2+3+4+5\right).\left(1.5.6\right)}{\left(1+2+3+4+5\right).\left(1.3.5\right)}\) = \(\dfrac{1.5.6}{1.3.5}\) = 2
\(A=\dfrac{15\left(1+2\cdot4+64\right)}{35+240+2240}\)
\(=\dfrac{15\cdot73}{2515}=\dfrac{15\cdot73}{5\cdot503}=\dfrac{3\cdot73}{503}=\dfrac{219}{503}>\dfrac{3}{8}\)
\(\frac{1.3.5+2.6.10+4.12.20}{1.5.7+2.10.14+4.20.28}\)
\(=\frac{3.5+2.3.2.5.2+4.3.4.5.4}{5.7+2.5.2.2.7+4.4.5.7.4}\)
\(=\frac{3.5.\left(1+2.2.2+4.4.4\right)}{5.7.\left(1+2.2.2+4.4.4\right)}\)
\(=\frac{3}{7}>\frac{3}{8}\)
\(2E=\frac{6}{1.3.5}+\frac{6}{3.5.7}+...+\frac{3}{13.15.17}\)
\(2E=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{13.15}-\frac{1}{15.17}\)
\(2E=\frac{1}{1.3}-\frac{1}{15.17}\)
\(2E=\frac{1}{15}-\frac{1}{255}\)
\(\Rightarrow2E=\frac{16}{255}\)
\(\Rightarrow E=\frac{8}{255}\)
a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)
\(A=\frac{1.5.6+1.5.6.2^3+1.5.6.3^3+1.5.6.4^3+1.5.6.5^3}{1.3.5+1.3.5.2^3+1.3.5.3^3+1.3.5.4^3+1.3.5.5^3}\)
\(=\frac{1.5.6.\left(1+2^3+3^3+4^3+5^3\right)}{1.3.5.\left(1+2^3+3^3+4^3+5^3\right)}\)
\(=\frac{1.5.6}{1.3.5}=\frac{1.5.3.2}{1.3.5}=2\)